the equation of motion for an object thrown from (0,0) at an angle θ with velocity v is
y(x) = -g/(2v2 cos2θ) x2 + xtanθ
the range (where y=0 again) is
r = v2 sin2θ/g
the maximum height reached is
h = v2 sin2θ/2g
So, we know that
h = 10
θ = 60°
10 = v2 (3/4)/(2*9.8)
10 = .038 v2
v2 = 263.16
v = 16.22
The range is twice the distance to the balcony, so the balcony is at half the range:
r = 16.22 sin(120)/9.8
= 29.21 * √3/2 / 9.8
= 2.58
so, he stood 1.29m from the house
Four masses are positioned at the corners of a rectangle, as indicated in the figure below (not to scale).
(a) Find the magnitude and direction of the net force acting on the 2.0 kg mass if x = 0.40 m and y = 0.12 m.
(b) How do your answers to part (a) change (if at all) if all sides of the rectangle are doubled in length?
The magnitude of the force will be unchanged.
The magnitude of the force will be reduced by a factor of two.
The magnitude of the force will be reduced by a factor of four.
The direction will remain unchanged.
The direction will shift clockwise.
The direction will shift counterclockwise.
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